1964 AHSME Problems/Problem 13

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Problem 13

A circle is inscribed in a triangle with side lengths $8, 13$, and $17$. Let the segments of the side of length $8$, made by a point of tangency, be $r$ and $s$, with $r<s$. What is the ratio $r:s$?

$\textbf{(A)}\ 1:3 \qquad \textbf{(B)}\ 2:5 \qquad \textbf{(C)}\ 1:2 \qquad \textbf{(D)}\ 2:3 \qquad \textbf{(E)}\ 3:4$

Solution

Label our triangle $ABC$ where that $AB=17$, $AC=13$, and $BC=8$. Let $L$, $J$, and $K$ be the tangency points of $BC$, $AC$, and $AB$ respectively. Let $AK=x$, which implies $AJ=x$. Thus $KB=BL=17-x$ and $JC=LC=13-x$.

Since $BL+LC=(17-x)+(13-x)=8$, $x=11$. Thus $r:s=CL:BL=13-x:17-x=2:6=1:3$, hence our answer is $\fbox{A}$.

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